GCSE Chemistry - How to Find the Volume of a Gas #28

Summary

This video explains how to calculate the volume of gases using a simple equation that multiplies the number of moles by 24, which works for all gases at room temperature and pressure. It covers examples like chlorine, oxygen, and water vapor, and discusses rearranging the equation to find moles from volume. The video also addresses finding gas volumes from chemical reactions using molar ratios, and highlights shortcuts when only converting between gas volumes. The method applies under standard conditions, emphasizing its use in exams.

Takeaways

• 📏 The equation for calculating the volume of a gas (in dm³) is simple: multiply the number of moles by 24.
• 🧪 The type of gas (e.g., chlorine, water vapor, oxygen) doesn't matter in this equation.
• 🔢 Example: 3.5 moles of chlorine gas takes up 84 dm³ of volume (3.5 x 24).
• 🔄 The equation can be rearranged to calculate moles if the volume is known (volume ÷ 24 = moles).
• 💧 Example: 60 dm³ of oxygen equals 2.5 moles of oxygen (60 ÷ 24 = 2.5).
• 💡 For mass calculations, use the formula mass ÷ Mr (molar mass) to find moles, then use the gas equation to find volume.
• 🚰 Example: 27 grams of water vapor with an Mr of 18 equals 1.5 moles, which occupies 36 dm³ of volume (1.5 x 24).
• 🧫 When reacting gases, the volume of the product depends on the limiting reactant, in this case, nitrogen.
• 🔗 Use molar ratios from the balanced equation to calculate product volumes (e.g., 18 dm³ nitrogen forms 36 dm³ ammonia).
• ⚠️ The 24 constant only works for gases at room temperature and pressure; changes in temperature or pressure would affect the gas volume.

Q & A

What is the basic equation used to calculate the volume of a gas?

The volume of a gas can be calculated by multiplying the number of moles of the gas by 24, as long as the gas is at room temperature and pressure. The result is the volume in decimeters cubed.

Does the type of gas affect the calculation of volume using this equation?

No, the type of gas does not affect the calculation. Whether it’s chlorine, water vapor, or oxygen, the same equation can be applied.

How would you calculate the volume of chlorine gas if you have 3.5 moles of it?

To find the volume of 3.5 moles of chlorine gas, you multiply the number of moles (3.5) by 24. The result is 84 decimeters cubed.

How can the equation be rearranged to find the number of moles from the volume of gas?

To find the number of moles from the volume, the equation can be rearranged to 'volume ÷ 24 = moles.' For example, if you have 60 dm³ of oxygen, you divide 60 by 24 to get 2.5 moles of oxygen.

How do you find the volume of a gas if you're given its mass instead of moles?

To find the volume from mass, first calculate the number of moles using the equation 'moles = mass ÷ relative formula mass (Mr).' Once you have the moles, multiply them by 24 to find the volume in decimeters cubed.

What is the process for finding the volume of water vapor if you are given 27 grams of it?

First, calculate the moles of water vapor by dividing the mass (27 grams) by the relative formula mass of water (18). This gives 1.5 moles. Then multiply 1.5 moles by 24 to get a volume of 36 dm³.

How do you calculate the volume of ammonia produced from 18 dm³ of nitrogen in a reaction with excess hydrogen?

First, calculate the moles of nitrogen by dividing 18 dm³ by 24, which gives 0.75 moles. Using the molar ratio of nitrogen to ammonia (1:2), you find that 0.75 moles of nitrogen will produce 1.5 moles of ammonia. Multiply 1.5 by 24 to get 36 dm³ of ammonia.

Can you simplify the process of calculating gas volumes when only gas volumes are involved?

Yes, if you're only converting between gas volumes, you can use the molar ratio directly without finding moles. For instance, if the ratio of nitrogen to ammonia is 1:2, you can multiply the volume of nitrogen by 2 to get the volume of ammonia.

How much hydrogen would react with 4 dm³ of nitrogen based on the molar ratio?

Since the molar ratio of nitrogen to hydrogen is 1:3, you multiply the nitrogen volume (4 dm³) by 3 to get 12 dm³ of hydrogen.

Does this equation with the number 24 work at any temperature and pressure?

No, this equation with the number 24 only works for gases at room temperature and pressure. Changing the temperature or pressure would alter the gas volume.

Outlines

00:00

🔬 Understanding Gas Volume and Moles Relationship

In this section, the video explains a basic equation that links the volume of a gas (in decimeters cubed) to the number of moles. The formula is simple: multiply the number of moles by 24 to get the volume. This method applies to any gas, whether it’s chlorine, water vapor, or oxygen. For instance, 3.5 moles of chlorine gas take up 84 decimeters cubed. The reverse calculation is also possible by dividing the volume by 24 to find the number of moles. For example, 60 decimeters cubed of oxygen would equal 2.5 moles.

05:02

📏 Converting Mass to Volume Using Two Equations

The video continues by addressing more complex scenarios where moles aren't directly provided. To calculate the volume of gas from mass, another equation is introduced that links mass, moles, and the relative formula mass (Mr). The example used is 27 grams of water vapor. First, you find the number of moles by dividing the mass by the Mr of water (H2O), which is 18. This gives 1.5 moles. Finally, multiplying by 24, the volume is found to be 36 decimeters cubed.

🧪 Calculating Volume of a Product from a Reactant in a Reaction

This part of the video shifts focus to calculating the volume of a product gas (ammonia) when given the volume of a reactant gas (nitrogen). Given that nitrogen is the limiting reagent, the moles of nitrogen are calculated by dividing its volume (18 decimeters cubed) by 24. Using the molar ratio between nitrogen and ammonia (1:2), 0.75 moles of nitrogen would produce 1.5 moles of ammonia, and multiplying by 24 gives a final ammonia volume of 36 decimeters cubed.

🔄 Shortcut for Volume Calculations Using Molar Ratios

The video introduces a useful shortcut for gas volume conversions: instead of calculating moles, you can directly use the molar ratio between gases to find the volume. For instance, to find the volume of ammonia produced from 18 decimeters cubed of nitrogen, the molar ratio of 1:2 is used to multiply the nitrogen volume by 2, resulting in 36 decimeters cubed of ammonia. This shortcut works because doubling the moles of gas also doubles its volume.

⚖️ Using Molar Ratios for Other Gas Reactions

Another example is presented, where 4 decimeters cubed of nitrogen is used to calculate the volume of hydrogen it would react with. Using the molar ratio of nitrogen to hydrogen (1:3), you multiply the nitrogen volume by 3 to find that it would react with 12 decimeters cubed of hydrogen. This method only applies when converting between gas volumes, not moles or mass.

🌡 Limitations of the Gas Volume Equation at Different Conditions

The final point covered in the video highlights that the gas volume equation using the number 24 only works at room temperature and pressure. Changing either temperature or pressure would alter the volume a gas occupies. However, for exam purposes, this simplified equation using 24 is the only one needed. The video wraps up by encouraging viewers to practice with these principles for exam success.

Keywords

💡Volume

Volume refers to the amount of space that a gas occupies, measured in decimeters cubed (dm³) in this video. It plays a central role in the video's explanation of gas laws, as the equation links volume to the number of moles of gas. For example, 3.5 moles of chlorine gas occupy 84 dm³ of space.

💡Moles

Moles measure the quantity of a substance, specifically the number of molecules or atoms. The video uses moles to explain the relationship between the amount of gas and its volume. For instance, the video calculates moles of oxygen from its volume (60 dm³ / 24 = 2.5 moles).

💡24

The number 24 in the video is a constant that represents the volume (in dm³) occupied by 1 mole of gas at room temperature and pressure. It is crucial in calculating the volume of any gas by multiplying it with the number of moles, or dividing volume by 24 to find moles.

💡Room temperature and pressure

Room temperature and pressure are the conditions under which the equation with the number 24 applies. This refers to standard atmospheric conditions, under which 1 mole of gas occupies 24 dm³. The video notes that changing these conditions would alter the gas volume.

💡Relative formula mass (Mr)

Relative formula mass (Mr) is the sum of the atomic masses of all atoms in a molecule. In the video, it is used to calculate moles from mass, such as when calculating the moles of water vapor from 27 grams of water by dividing its mass by the Mr of H2O (18).

💡Limiting reagent

A limiting reagent is the reactant that limits the amount of product formed in a reaction. In the video, nitrogen is the limiting reagent when reacting with excess hydrogen to form ammonia, meaning the quantity of nitrogen determines how much ammonia is produced.

💡Molar ratio

Molar ratio is the ratio of moles of reactants to products in a chemical reaction, determined by the coefficients in a balanced equation. The video uses a molar ratio of 1:2 between nitrogen and ammonia to calculate how much ammonia is formed from a given volume of nitrogen.

💡Gas volumes

Gas volumes refer to the measurable amount of gas produced or required in a reaction, expressed in dm³. The video shows how to calculate gas volumes using moles and molar ratios, such as when determining that 18 dm³ of nitrogen will produce 36 dm³ of ammonia.

💡Excess reactant

An excess reactant is the substance that is available in greater quantity than necessary for a reaction. In the video, hydrogen is the excess reactant when reacting with nitrogen to form ammonia, meaning the amount of hydrogen does not limit the reaction outcome.

💡Mass

Mass refers to the amount of matter in a substance, measured in grams. In the video, mass is used to calculate moles when the number of moles is not provided, as seen in the calculation of moles from the 27 grams of water vapor using its relative formula mass.

Highlights

Introduction of an equation that links the volume of a gas in decimeters cubed to the number of moles of gas.

The equation is simple because it works for any type of gas, regardless of its identity.

To calculate the volume of a gas, multiply the number of moles by 24 to get the volume in decimeters cubed.

Example given: 3.5 moles of chlorine gas would occupy 84 decimeters cubed.

Rearranging the equation allows for calculating moles when given a volume, such as 60 decimeters cubed of oxygen equating to 2.5 moles.

To find the volume of a gas given its mass, first calculate the moles using mass and relative formula mass.

Example of water vapor: 27 grams of water vapor with a relative formula mass of 18 gives 1.5 moles.

Using the original equation, 1.5 moles of water vapor occupies 36 decimeters cubed.

In a reaction where nitrogen reacts with excess hydrogen, the limiting reagent is nitrogen.

For 18 decimeters cubed of nitrogen, there are 0.75 moles.

Using the molar ratio of nitrogen to ammonia (1:2), 1.5 moles of ammonia is produced.

This results in 36 decimeters cubed of ammonia from the 18 decimeters cubed of nitrogen.

When only converting between gas volumes, it is unnecessary to calculate moles, as long as the molar ratio is known.

Example: 4 decimeters cubed of nitrogen reacts with 12 decimeters cubed of hydrogen, based on the molar ratio (1:3).

The equation with the number 24 only works for gases at room temperature and pressure.

Transcripts

in today's video we're going to be

focusing on this equation

which links the volume of a gas in

decimeters cubed

to how many moles of that gas we have

it's actually a really simple equation

because it doesn't matter which type of

gas you have

like whether you have chlorine or water

vapor or oxygen

all you have to do is multiply the

number of moles that you have

by the number 24

and that will give you the volume that

the gas takes up

measured in decimeters cubed

for example if we had 3.5 moles of

chlorine gas

then we would just do 3.5

times 24

to find that it took up 84 decimeters

cubed

we can also use the equation the other

way around

for example if we had 60 decimeters

cubed of oxygen

and we wanted to know how many moles

that was

we'd just rearrange the equation

to volume over 24 equals moles

and then do 60 divided by 24

which gives us 2.5 moles of oxygen

it can get a bit trickier if we're not

given the moles

for example if we had 27 grams of water

vapor

how would we find its volume

well we can see from our equation that

to find the volume we first need to find

the moles

so we're going to have to use this other

equation first

which links mass

moles and relative formula mass

already know that the mass is 27 grams

so we just need the mr

which for water which is h2o

would be 16 for the oxygen

plus 2 times 1 for the hydrogens

so eighteen

then we can divide the mass of 27 grams

by the mr of 18

to find that we must have 1.5 moles of

water vapor

then lastly all we need to do is go back

to our original equation

take our 1.5 moles

and multiply it by 24

to find that the water vapor would have

a volume of 36 decimeters cubed

another thing you might be asked to do

in the exam

is to find the volume of a product when

you're given the volume of a reactant

for example in this equation here

what volume of ammonia would be produced

if we reacted 18 decimeters cubed of

nitrogen with excess hydrogen

the first thing to notice is that the

hydrogen is in excess

which means that the nitrogen must be

the limiting reagent

and so the quantity of ammonia produced

is going to depend entirely on how much

nitrogen we have

next the normal thing to do in this sort

of question would be to find out how

many moles of nitrogen we have

so we take our volume of 18 decimeters

cubed

and divide it by 24

which tells us that we have

0.75 moles of nitrogen

and then we can use the molar ratio to

find out how many moles of ammonia that

will form

remember we find the molar ratio by

comparing these big numbers in front of

the chemical symbols

so an imaginary one for nitrogen because

when there's no number it just means

there's an unwritten one

and a two for ammonia

so the ratio is one to two

which tells us that for every mole of

nitrogen we have

we're going to make two moles of ammonia

so as we have 0.75 moles of nitrogen

we must have 0.75 times 2 moles of

ammonia

so 1.5 moles of ammonia

then finally we just go back to our

original equation

and multiply the 1.5 moles of ammonia by

24

to find that we'll make 36 decimeters

cubed of ammonia

now this whole idea of finding the moles

and then using molar ratios

is what you would normally have to do in

a question like this

however if you're only converting

between gas volumes like we are here

then you don't actually have to do any

of this stuff

all you need to do is look at the modal

ratio

which is one to two

and so you can multiply the original

volume of nitrogen which is 18

decimeters cubed

by two

to find that you'd make 36 decimeters

cubed of ammonia

and the reason we can do this is because

if we have twice as many moles of a gas

it's going to take up twice the volume

to see the zim practice let's change the

question a bit

if we start with four decimeters cubed

of nitrogen

how much hydrogen would it react with

because we're only converting gas

volumes we can just look at the molar

ratio between nitrogen and hydrogen

which is one two three

so then we just take the nitrogen's

volume of four decimeters cubed

and multiply it by 3

to find that it would react with 12

decimeters cubed of hydrogen

the very last thing i want to mention

is that this equation with the number 24

only works for gases at room temperature

and pressure

if we change the temperature or the

pressure

then it would change the number

as gases occupy different volumes at

different temperatures or pressures

but in the exams this equation with the

number 24 is the only one that you'll

need to use

anyway that's everything for this video

so hope you found it useful and we'll

see you again soon